JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Kinematics

The trajectory of a projectile near the surface of the earth is given as

y = 2 x - 9 x^{2}

. If it were launched at an angle

θ_{0}

with speed

v_{0}

, then

(g = 10 m s^{- 2})

Answer: (A)

Solution

The projectile trajectory has the standard shape

y = x tan θ_{0} - \frac{g}{2 u ^{2} cos ^{2} θ _{0}} x^{2}

. Matching it to

y = 2 x - 9 x^{2}

, the linear coefficient gives

tan θ_{0} = 2

, so

cos θ_{0} = \frac{1}{5}

and

θ_{0} = cos^{- 1} \frac{1}{5}

. The quadratic coefficient gives

\frac{g}{2 u ^{2} cos ^{2} θ _{0}} = 9

; with

g = 10

and

cos^{2} θ_{0} = \frac{1}{5}

this is

\frac{10}{2 u ^{2} \cdot \frac{1}{5}} = 9

, so

u^{2} = \frac{25}{9}

and

u = \frac{5}{3} m/s

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option